Our school's girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, and Anna.  In how many ways can we choose 6 starters if at most one of the triplets is in the starting lineup?
Solution: We can add together the number of lineups with one triplet and with no triplets.   The number of lineups with no triplets is $\binom{11}{6} = 462$, since we must choose 6 starters from the 11 remaining players.  When one triplet is in the lineup, there are $3\cdot \binom{11}{5} = 1386$ options.  So the total number of lineups with at most one triplet is $1386 + 462 = \boxed{1848}$.